Problem: What is the extraneous solution to these equations? $\dfrac{x^2 - x}{x - 9} = \dfrac{11x - 27}{x - 9}$
Multiply both sides by $x - 9$ $ \dfrac{x^2 - x}{x - 9} (x - 9) = \dfrac{11x - 27}{x - 9} (x - 9)$ $ x^2 - x = 11x - 27$ Subtract $11x - 27$ from both sides: $ x^2 - x - (11x - 27) = 11x - 27 - (11x - 27)$ $ x^2 - x - 11x + 27 = 0$ $ x^2 - 12x + 27 = 0$ Factor the expression: $ (x - 3)(x - 9) = 0$ Therefore $x = 3$ or $x = 9$ At $x = 9$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 9$, it is an extraneous solution.